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Abbreviations

1- TTDh Heater Terminal Temperature Difference. 2- TTDc = Condenser Terminal Temperature Difference.3- Tloss = temperature difference between brine and condensing vapor temperature.

To Simplify the Single Stage Flash in Mathematical Analysis, the following assumption must be considered :

1-Specific heat at constant pressure(Cp) = 4.18 kJ/kg C.

2- The overall heat transfer coefficient Uh = Uc = 2 KW/m2 c

3- Neglect the Steam Sub-cooling and super heating

4-Neglect all heat losses

5- Xp zero ( Salt Free)

1- Mf = Mb + Md

2- Xf Mf = Xb Mb

3,4) Ms s = Mf cp (To T1 ) = Uh Ah (LMTD)h

Where :

s : latent heat of steam

Uh: heater overall heat transfer coefficient (KW/m2 C)

Ah : Area of heater (m2)

(LMTD)h : Logarithmic Mean Temperature Difference of heater

5,6) Md v = (Mcw + Mf ) cp ( T1 Tcw )= Mf cp ( To Tb) = Uc Ac (LMTD)c

Where :

v : latent heat of Vapor

Uc: overall heat transfer coefficient (KW/m2 c)

Ac : Area of condenser (m2)

(LMTD)c : Logarithmic Mean Temperature Difference of condenser

Where :

(LMTD)h : Logarithmic Mean Temperature Difference of heater

To: Top brine temperature

Ts: Steam temperature

(To T1)7) (LMTD)h =

ln (Ts T1 )(Ts To )

Where :

(LMTD)c : Logarithmic Mean Temperature Difference of condenser

Tcw: cold water temperature

Tv: vapor temperature

(T1 Tcw)8) (LMTD)c =

ln (Tv Tcw)(Tv T1 )

Where :

PR : Performance Ratio

s : steam latent heat

v : vapor latent heat

Md s Mf cp (To Tb ) s (To Tb ) 9) PR = = =

Ms v Mf cp (To T1 ) v (To T1 )

10) Tst = To - Tb

11) To T1 = Tst + T loss+ TTDcWhere :

TTDc : Condenser Terminal Temperature Difference

Tloss : Thermodynamic losses

Tst : stage temperature drop

C

Given

To : Top brine temperature = 90c Tb: Temperature of reject brine= 40 C

Tcw : cold water temperature = 30cTs : Steam temperature = 100c

s : steam latent heat = 2256 kJ/kg (at 100 C)

v : vapor latent heat = 2412.5 kJ/kg (at 38 C)

Xf : The salinity of intake seawater = 42,000 ppm

TTDc : The condenser terminal temperature difference = 3c

T loss : Thermodynamic loss = 2c

Required

Performance Ratio

Salinity of brine

Area of heater (m2)

Area of condenser (m2)

PR

Mf/ Md

Mcw / Md

Xb

Ah

Qin / Md

Ac

specific flow rate of the cooling water

specific flow rates of feed seawater

Solution

Tst = To - Tb = 90 - 40 = 50CTTDh = Ts To= 100 - 90 = 10CTv = Tb - Tloss = 40 - 2 = 38Cv = 2412.5 kJ/kg (at 38

oC)s = 2256 kJ/kg (at 100

oC)

(s)(Tst )

PR =

(Tst + Tloss +TTDc) (v)

= ((2256) (50)) /((50+2+3)(2412.5))= 0.85 kg distillate/kg steam

Solution

Md v = Mf Cp TstMf/Md = v /(Cp (Tst ))

= 2412.5/((4.18)(50)) = 11.54 kg feed seawater/kg distillate

Md v = (Mf+Mcw) Cp (To- Tst - Tloss-TTDc-Tcw)Mcw/Md = v(Cp (To- Tst -Tloss-TTDc-Tcw)) - (Mf/Md)

= (2412.5)/((4.18)(90-50-2-3-30)) - 11.54 = 103.9 kg cooling seawater/kg distillate

Mb = 11.54 - 1 = 10.54 kg brine/kg distillateXb = Xf Mf/Mb = (42000) (11.54)/(10.54) = 45984.8 ppm

Solution

((Ms) s))

Ah/Md =

((Md)(Uh)(LMTD)h)

= 2256/((0.85)(2)(29.4))

= 45.1 m2/(kg/s)

(LMTD)h = Tst + Tloss + TTDc

ln (( TTDh + Tst + Tloss + TTDc) / TTDh )

= (50+2+3) / ln((10+50+3+2)/(10))

= 29.4 oC

Solution

Ac/Md = v /((Uc)(LMTD)c)

= 2412.5/((2)(17.4))= 69.3 m2/(kg/s)

Qin /Md = Ms .s/Md

= 50/ln((50+3)/3)= 17.4 C

Tstln((Tst +TTDc)/(TTDc))

= 2256/.85 = 2654.11 kJ/ kg distilled water

(LMTD)c =

This method is a great improvement in the single stage flush

The same process as single stage flush. The process is repeated several time and It recommend with the

same temperature difference

The target is to decrease the amount of energy stored in the steam so the process dont need to be cooled.

So the maximum efficiency increased.

1) Md v = Mf cp (To Tn )= Mf cp n. Tst

Where :v: latent heat of vapor @ ( T = (To + Tb )/2 )Md : MdiTo Tb : n Tst

As :n :is the number of stagesTst :is the average temperature difference of stage (recommended to be the same difference).

Required

Performance Ratio

Salinity of brine

Area of heater (m2)

Area of condenser (m2)

PR

Mp / Md

Xb

Ah / Md

Ac / Md

specific flow rates of feed seawater

Tst = (To Tn ) / 23 = (90-40)/23 = 2.174 C

Taverage = (To + Tn ) / 2 = (90+40)/2 = 65C

v = 2346 kJ/kg

(n Tst )sPR =

(Tst + Tloss + TTDc ) v= (23)(2.174)(2256)/((2.174+2+3)(2346.5))

= 6.7 kg distillate water/kg steam

Solution

Solution

Solution

Xb = Mf .Xf /Mb = (42000) (11.22)/(10.22) = 46,126 ppm

Mf/Md = v/(Cp(To - Tb))

= 2346.5/((4.18)(90-40))= 11.22 kg intake seawater/kg distillate

Mb/Md = 11.22 - 1 = 10.22 kg/kg distillate

The specific feed flow rate is

Solution

= (2.174+2+3)/ln((10+2.174+2+3)/(10))

= 13.27 C

Ah / Md = Ms s / (Md Uh (LMTD)h

= 2256/((6.7)(2)(13.27))

= 12.7 m2/(kg/s)

(LMTD)h = Tst + Tloss + TTDc

ln (( TTDh + Tst + Tloss + TTDc) / TTDh )

Solution

(LMTD)c = (T st ) / ln (T st + TTDc )/(TTDc))

= 2.174/ln((2.174+3)/3) = 3.98 C

Ac / Md = Mf Cp Tst / (Uc (LMTD)c )=(11.22)(4.18)(2.174)/((2)(3.98))

= 12.8 m2/(kg/s distilled water)

Total area = Ah + n Ac = 307.1 m2

Conclusion:. Single stage once through type is more efficient and more

economical in energy consumption.. Single stage once through type is more suitable for

environment.. Single stage once through type need more area to be

executed.

Until now we have discussed two different types of waterdistillation (the single stage flash, and the single through multistage flashing) accompanied with a proper Thermodynamicsanalysis of the systems.

In this section we will discuss a third type of multi stageflashing which is MSF with brine mixing, and by the end ofreading this lecture you should:

1. Know the theory of operation of this type2. Spot the advantages and disadvantages of this type.3. Understand the thermodynamics analysis of this type.4. Compare between the three types with respect to (PR)

and total Area.

productSteam in

Steam Out(Condensate)

Reject

Feed water

Bri

ne

to

be

m

ixe

d

(schematic diagram of MSF with brine mixing)

According to previous figure:

It is a basic MSF except that a portion of a mass of a value equalto (Mr Mf) was taken from the exiting brine with temperature of(Tb) to be mixed with the entering feed water having the mass(Mf) & temperature (Tcw) raising its temperature to (Tr) butincreasing its salinity as well.

After the mixing process, the steps are the same as any MSFdistillation process.

What is its idea?It is simply based on the idea of making a use of the brine

outcome by recirculating a portion of it in the MSF cycle , thusmaking use of its thermal energy instead of throwing it away.

Why do we insert a saline solution back into the system ?In this type of desalination it is acceptable to do so -as long as

we are operating within the predetermined limitations (salinity up to70,000 ppm)- because we are using thermal energy to extract freshwater by means of evaporation and condensation.

What happens if we exceeded the 70,000 ppm limit?This will cause the following problems in the cycle:

1. Will cause excessive fouling and scaling throughout the system.2. Decreases the overall efficiency.3. The resulting reject could have huge impacts on the environment, which

would require more handling.

What is the purpose of using brine recirculation (Adv.)?1. Decreasing the feed water flow rate which decreases the amount of the pre-

treatment required for the feed water.2. Improving the thermal efficiency of the cycle by recirculating the brine which

has higher thermal energy than feed water, instead of throwing it away.3. Decreasing the amount of brine water exiting the system.

What are the main disadvantages of using brine recirculation?1. Due to excessive salinity the system requires frequent cleaning and maintenance in order to prevent scaling.2. Requires larger area.

1. By applying mass balance to the whole system we get,

M = Mb + Md .. . Equation ( 1 )

2. By applying salinity balance on the whole system we get,

M . X = Mb . Xb + Md . Xd

( Xd ) assumed to be zero, due to its very small value Therefore,

M . X = Mb . Xb .. . Equation ( 2 )

From ( 1 ) and (2 ):

X (Mb + Md ) = Mb . XbMd . X = Mb ( Xb - X )

XMb = . Md .. . Equation (3 )

Xb - X Where,

Mb is the mass flow rate of brine.Md is the mass flow of distilled water.X , Xb are the salinity of both feed and distilled water respectively.

(Figure 2)

4,5. By applying energy balance on the heater ( figure 2 )we get,

Ms . s = Mr . Cp . (To - T1 )

Md . v = Mr . Cp . (To - Tb)

. Equation ( 4 )

. Equation ( 5 )

6. And from previous lecture we could deduce that,

Md (To - Tb ) . sPR = = .. . Equation ( 6 )

Ms (To - T1 ) . v

Note: the performance ratio may be considered as a measure of efficiency.

. Equation (7)

Figure(8.4)

Step 1: Analyzing the given information and requirements

Given:

Required to find:

. From equation ( 8.3 ), get the mass of exiting brine :

( X ) * ( Md )Mb=

( Xb Xf )( 42000 ) * ( 1 )

Mb= ( 70000 42000 )

Mb = 1.5 Kg/S

. And From equation ( 8.1 ), We could get the mass of entering water:Mf = Mb + Md

Mf = 1+ 1.5 = 2.5 Kg/S

. Knowing the value of PR ( from equation 8.4 ), get the values ofMs & T1 , or do energy balance for the heater to get T1:

Md 1PR = 8 =

Ms Ms

Therefore, Ms = 0.125 Kg/S

PR=Md/Ms=8, and Md=1 kg/s then Ms=0.125 kg/s

For the heater we have ,

( Ms ) * ( s ) = ( Mr ) * ( Cp ) * ( To T1 )

( 0.125 ) * ( 2200 ) = ( 17.942 ) * ( 4.18 ) * ( 90 T1 )

Which gives out that T1 = 86.66 cwe had deduced that,

Md (To - Tb ) . sPR = = =8 .. . Equation ( 6 )

Ms (To - T1 ) . v

Then Tb = 60.8 deg C

Also, To-N.Tst=Tb

Then Tb = 90 23 * Tst then Tst =1.3 deg C

. And we know that,( Md ) * ( v )

Mr * cp =(To - Tb )

. Then,( 1 ) * ( 2400 )

Mr * 4.18 =( 90 60.8 )

Therefore, Mr = 17.942 Kg/S

( Md ) * ( Xf ) + ( Mr Mf ) * Xb = ( Mr )* ( Xr )

Therefore, Xr = 66098 PPM

.

. LMTDh =

. LMTDh = =12.93C

. LMTDc = =

then, LMTDc= 3.65C

last step area calculationAcondenser*Uc*LMTDC=Mr*CP*TstageAcondenser*2*3.65=17.942*4.18*1.3then

Acondencer= 14 m2

Aheater*Uh*LMTDh=Ms*sAheater*2*12.93=0.125*2200then

Aheater=10.634 m2

Type SSFMSF once through

MSF with brine

mixing

PR 0.67 6.7 8

Total Area (m^2)

105 302 332

This table has a great significance, since it means that we have improved the PR but it with accompanied by size issues (increased size ).

In this type we utilized the useful thermal energy going out with the reject, thus maximizing the PR values, and increasing the recovered water amount.

The increase in the PR values is accompanied with an increase in total area.

Frequent maintenance and cleaning is a must with this type of distillation to prevent fouling or scaling that may occur due to the high percentage of salinity.

The maximum allowable amount of salinity to be recirculated is about 70,000 ppm.

If the maximum value was exceeded, scaling and fouling will take place in the system, as well as an increase in the environmental impact of the reject.

5858

Item Details

No. of Students per topic Three Students maximum

Output Format Power Point + videos

Equations editing To be written on equation editor; not photo

Dead line 2nd & 9th of May, 2014

Presentation 15-20 minutes maximum

Deliverable form CD including the presentation

Topic 1 Using CFD in Desalination

1. Finite difference and finite element

2. Main Boiling Equations and solution assumptions

3. Main Flush Equations and solution assumptions

4. Main Flush Equations and solution assumptions

5. Main salt separation Equations and solution assumptions

5959

Item Details

Topic 2 Methods of Humidification and Dehumidification Desalination

1. Open Air Open Water Systems

2. Open Air Closed Water Systems

3. Closed Air Open Water Systems

Topic 3 Hybrid Solar Energy and Desalination

1. Available Arrangements

2. Relevant calculations

3. Case Studies with its calculations (minimum two cases)

Topic 4 Hybrid Wind Energy and Desalination

1. Available Arrangements

2. Relevant calculations

3. Case Studies with its calculations (minimum two cases)

6060

Item Details

Topic 5 Hybrid Geothermal Energy and Desalination

1. Available Arrangements

2. Relevant calculations

3. Case Studies with its calculations (minimum two cases)

Topic 6 Arrangements of RO Membrane

1. Available practical membranes specification with catalog cuts

2. Arrangements of membranes (Series and Parallel)

3. Case Studies with its calculations (minimum two cases)

Topic 7 Full Design of MSF Plant

1. Practical Arrangements (not schematic)

2. Relevant calculations (Heat exchanger Fuel Steam -)

3. Case Studies with its calculations (minimum two cases)

6161

Topic 1 Asmaa Ramadan El Sayed

Ibrahim Gad El Haq

Topic 2 Ahmed Mahmed Osman

Wael Wadee

Mohamed Fouad

Topic 3 Ahmed Tarif

Mohamed Bashir

Waleed Mohamed

Topic 4 ---------------------

Topic 5 ---------------------

Topic 6 Ashraf Hussein

Mohamed Tarek

Mohamed Hussein

Topic 7 Mohamed Essam

Mohamed Hamdy

Aya Ahmed

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